TRANSIENT ANALYSIS OF LAYERED COMPOSITE PLATES ACCOUNTING FOR TRANSVERSE SHEAR STRAINS AND VON KARMAN STRAINS by DANIEL JOSEPH MOOK THESIS submitted to the Faculty of the Virginia Polytechnic Institute and State University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE in ENGINEERING MECHANICS APPROVED: J. N. REDDY DANIEL FREDERICK WAYNE W. STINCHCOMB June, 1982 Blacksburg, Virginia
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
TRANSIENT ANALYSIS OF LAYERED COMPOSITE PLATES ACCOUNTING FOR TRANSVERSE SHEAR STRAINS AND VON KARMAN STRAINS
by
DANIEL JOSEPH MOOK
THESIS submitted to the Faculty of the
Virginia Polytechnic Institute and State University
in partial fulfillment of the requirements for the degree of
MASTER OF SCIENCE
in
ENGINEERING MECHANICS
APPROVED:
J. N. REDDY
DANIEL FREDERICK WAYNE W. STINCHCOMB
June, 1982 Blacksburg, Virginia
ACKNOWLEDGEMENTS
Dr. J. N. Reddy, Professor ESM, Virginia Tech, made this work
possible. His expert advice and encouragement were indispensable to
the completion of this project, and will be forever appreciated. Dr.
Reddy also provided a large part of the computer code used in the
analysis, saving the author the typically frustrating task of debugging
a large portion of the program.
would also Ii ke to express my appreciation for the helpful com-
ments of Dr. Daniel Frederick and Dr. Wayne Stinchcomb, and to thank
them for serving on my thesis committee.
Karen Mook, wife of the beleaguered author, not only provided
many conveniences during the harrowing ordeal, but also contributed
several of the figures. In addition, she tediously served as a sounding
board for many versions of the text which appears herein. My love
goes with my respect and appreciation for her contributions.
It is a pleasure to acknowledge the skillful typing of the equa-
tions and the figure captions by Sharon Larkins, my friend and neigh-
bor.
This work was predominantly funded by a research grant from
the Structural Mechanics Section of the Air Force Office of Scientific
Research (Grant AFOSR-81-0142). The support is gratefully acknowl-
+ [6\JJ MG], + [6\jJ Mz], + 50 Q:i.dxdy ·y x y y x ~
(3.1.3)
20
Using the divergence theorem, some of the volume integrals in eqn.
(3.2.3) may be converted to surface integrals. Specifically,
= J: 5 ( unNn + 6usNns)ds n
(o~ M + ow M )ds n n · s ns
where C , C , and C are shown in Fig. 3. n q m
8wqds
(3.1.4)
The right hand side of eqn. (3.1.4) clearly indicates the boundary
conditions. On each face of the four-sided (quadrilateral) element,
specify exactly one of each of the following pairs:
u or N n n
u or N ns ns
w or q (3.1.5)
1~ or M n n
') or M ns ns
21
where the subscript · n' indicates normal and · s' indicates tangential.
3 .3 THE FINITE ELEMENT PROGRAM
Time and space constraints prevent describing the finite-element
method (FEM) in any more detail than a rudimentary discussion. In or-
der to provide the most useful information in the limited space availa-
ble, the structure and execution of the computer program used in the
analysis is outlined in appendices A, B, and C. The rest of this chap-
ter outlines the procedures used in the program and discusses certain
areas where engineering judgement is required.
3 .3. 1 BRIEF OUTLINE OF THE FEM
As used in this analysis, the FEM consists of the following general
steps
1. Write the governing equations for the medium and then state the
minimum potential (strain) energy using standard variational
p raced u res .
2. Discretize the domain into 'finite elements'.
3. Select approximation functions (often called the 'shape func-
tions') to be used to approximate the solution in each element.
4. Derive the element governing equations using the equations from
step 1 along with the approximation functions from step 3.
5. Assemble the element governing equations from step 4 into glob-
al governing equations. A set of simultaneous equations is ob-
tained which may be written in the form
[M]{~} + [K]{~} = F(t) (3.3.1)
22
where [M] is the 'mass matrix', [K] is the 'stiffness matrix',
{.6.} is the vector of node-point displacements, {6} is the vector
of node-point accelerations, and F(t) is the node-point force
vector. This is a 'displacement formulation·.
6. Reformulate the equations using the time-marching scheme out-
lined in the next section. This produces a set of simultaneous
equations which may be written
[K 1 ]{6} = F1 (t) (3.3.2)
7. Apply the boundary conditions to eqn. (3.3.2).
8. Solve for the unknown displacements in eqn. (3.3.2).
9. Solve for the so-called secondary quantities; e.g., use the
strain-displacement relations to find the strains, then use the
stress-strain relations to find the stresses.
3 .4 THE TIME-MARCHING SCHEME
Eqns. (3.3. 1) are valid at any particular time t. In order to in-
tegrate eqns. (3.3. 1), Newmark's direct integration technique is em-
ployed. Using this technique, the displacements and accelerations are
found at the discrete times to = 0, ti = ot, tz = 2ot, ... , t = not. The n . velocities {0.} and displacements{..:..} are found at time step k•l (i.e., t
= (k•1) ot) by the following assumed expressions
{. \. 6.J k+ 1
(3.4.1)
23
where the subscripts refer to the quantity at that time step and Ci and
$ are constants used to obtain accuracy and stability. ~l is determined
from 3 by the expression
(3.4.2)
For linear analysis, the choice a = t and 3 = ~ has been shown to
be unconditionally stable. No similar results for nonlinear analysis have
been reported in the literature.
A brief investigation of the effect of varying ct. and 3 in the
present nonlinear analysis was carried out by running the same problem
with several different values for the two parameters. The results are
shown in Fig. (4).
In Fig. (4), the effect of J. and 3 on stability is shown. It is
clear that for ::l = 0.1, 0.2, and 0.3, the solution becomes unstable be-
fore the peak deflection is reached. For ::_t = 0.4, the solution becomes
unstable near the end of the first deflection cycle. For a = 0.5, the
solution maintains stability throughout the deflection cycle. In all of
the results reported later, a value of J. = 0.5 was used and assumed to
be both stable and accurate.
The size of the time-step used in the time-marching equations is
also important for stability and accuracy. Too small a time-step will
add to the cost, perhaps significantly, but too large a time-step may
give unstable or inaccurate answers. At any given discrete time during
24
EFFECT OF T~·!O-LRYER c:::.; - :>· v ~it"'i :::i-1L1, LJ;JL., S 3 , S :) • P L A T E t4 ! T H A / f-! ~ 1 0
Q8P.Fl=50. E11E2=25
z l 0 t--1 I 1--
('\t u I I ~w
Ill I
-"' LL Lt.J 0 t\J:-
1
l 2 TTfvlC I l. I IL.
Figure 4 Effect of a on the Stability of the Solution.
The data for these cases is given in eqn. (4.4.1),
except for the variation of ~. Note that as ~
increases, the solution maintains stability longer
through the cycle. For a= 0.5. the solution is
stable throughout the cycle.
25
the time-marching algorithm, the program must iterate to find the nonli-
near solution. Thus, the total number of iterations required to inte-
grate the equations of motion for a specific problem is the sum of the
nonlinear iterations required for each discrete time during the time
march. If the time-step is too small, the number of total iterations may
be high even though the number of iterations at each discrete time may
be low. Increasing the size of the time-step may reduce the total num-
ber of iterations by reducing the number of discrete times at which a
solution is sought. However, if the time-step is too large, the program
may require enough extra iterations to converge at each discrete time
that the total number of iterations will be higher than the total for a
smaller time-step. Therefore, the size of the time-step may have a pro-
found effect on the efficiency (measured by the total number of itera-
tions) of the program.
Time-steps used in this analysis were determined in the following
manner. The solution was found using an initial time-increment value
of (5t) 0 • The solution was then calculated using a smaller value for
5to and the solutions from the two different time-increment values were
compared. If necessary, smaller and smaller time-increments were used
until the solutions converged. When the solutions were essentially iden-
tical for two different time-increment values, stability and accuracy
were assumed. Every result reported herein was obtained in this fa-
shion and is assumed to represent a stable and accurate solution.
Comparisons with well-known solutions for certain special cases are re-
ported in Chapter 4 and indicate that this method is indeed accurate.
26
3 .5 SPECIAL CONSIDERATIONS
The following observations concerning the actual programming are
presented without pretense to a complete explanation.
Discretization is often the most subjective phase of any finite-ele-
ment analysis. In this analysis, each node was allowed five degrees-of-
freedom (D.0.F.); u,v,w, ljJ, :JJ • Considering eqn. (3.3.1), the fi-x y
nite-element method eventually reduces to the solution of a system of
simultaneous equations, of the order of the total number of D. 0. F.
Since every node adds five to the order of the system, there is an ob-
vious savings (which may be astronomical!) to be gained by using as
few nodes as possible. On the other hand, having too few nodes may
adversely affect accuracy. In fact, since the program iterates to find a
solution within a user-supplied error bound, too few nodes may lead to
a large enough increase in the number of iterations to offset or even
exceed the savings of a small-order system. For the purpose of this
analysis, the element break-up shown in Figure (Sa) was used. A finer
mesh is shown in Figure (Sb). The mesh in Figure (Sb) cost approxi-
mately 10 times as much to execute as the mesh in Figure (Sa), yet the
results were indistinguishable.
The elements of the mass [M] and stiffness [K] matrices are giv-
en in Appendix A. The derivation involves rather lengthy manipula-
tions and is not included.
The five boundary conditions for any particular node are given in
eqn. (3.1.5). In Figure (Ga), the boundary conditions are shown for
27
(a)
I I l I
(b)
Figure 5 Finite Element Mesh.
For the test cases used, the mesh in
(a) produced results identical to the
mesh in (b) but at approximately 1/10
the cost.
28
the mesh used in this analysis. An alternate formulation of boundary
conditions is also possible and is shown in Figure (6b). The alternate
form produces a significant difference in both the magnitude and timing
of the response, as shown in Figure (7). The boundary conditions in
Figure (6a) are generally accepted as correct, and a comparision of re-
sults obtained using these boundary conditions with results obtained
from an exact solution for a linear homogeneous plate shows excellent
agreement (see Figure 8).
A flow chart of the program is included in Appendix 8, and a
listing of the program is provided in Appendix C.
(a) v = 0 }
!./) = 0 y
(b) u = 0 }
'Px = o
y
I (u -
(v = 0,
29
v
i w
J . ·j; 1..: x
t 0, ~ y = 0)
w = 0, i~ = 0) I x ...
( v = 0 ' 1JJ = 0) ' ·y
= 0
= 0
= 0
u = 0
~ \1 = 0
;j; = 0 y
x
------x
Figure 6 Alternate Forms of the Boundary Conditions.
'Results are given in Figure (7).
z
.__ tot u ~w
)II
_J LL w 0
-
30
EFFECT OF BOUNDARY CONDITIONS - TWO FORMS IN FIG. 6 TWO-LAIER CROSS-P~ I, UDL, SS, SQ. PLATE WITH A/H=lO
QSAR=SO, El/E2=25
I I 1 2 3
TIME xl 0·3
figure 7 Comparison of Results Using Different Boundary Conditions.
Note that (a) and (b) from Figure 6 give markedly different
results. The conditions in Figure (6a) are generally
regarded as correct.
4.1 COMMENTS
Chapter IV
VERIFICATION ANO RESULTS
In this chapter, the accuracy of the program is established and
then certain parametric effects are investigated. The accuracy will be
demonstrated by comparing results obtained using this program with
those available in the literature. Excellent agreement is evident.
Next, the effect of the transverse shear constant and the rotatory iner-
tia is illustrated for specific data cases (the effect is different for
every data case). Finally, the effects of load magnitude, material or-
thotropy, plate aspect ratio, plate thickness, and lamination scheme are
reported.
4.2 VERIFICATION
The program accuracy was initially checked against linear isotrop-
ic solutions. In Figure (9), the results from this program are plotted
along with a classical thin-plate theory solution due to Volterra and Za-
chmanoglou (11) and an analytical thick-plate solution due to Reismann
and Lee ( 12) .
The data for this program was as follows:
a = 4, b = 2, h = 0.2, ~ = 1 ' ot=o.1, \) = 0.3,
E = 1 , q 0 = 10 (4.2.1)
31
I ::J 2.0
1. 5
1. 0
0.5
o.
4..1 c: Cl) 8 0 e 20 ell t:: .....
"'O c: \lJ 15 .0
.... CJ .i.i c: >. ill l:Z u 10
"Cl Cl) N ....
r-4
"' c: 5 0 ..... Ul c: C,) s .....
"'O c 0 :z
32
Riesmann and
0 Present Th i,:k-r l.:ite theory
' ' ' ' Classical ' \ Plate theory\ \
' \ b3 \ I w = w Eha/q \
0 ' t tr./bq ' ' 0 ... .... ....
1. 2. 3. 4. s. 6. 7. 8. Nond imensionalized time, t
(a) Deflection versus ti~e
Riesmann and Le~O"'O\
(1969)/ \ 0 Present • Thick-plate
I__ ~
I I
' '
2
-'.-1 v
.... , '-.._,..-a ' ,...... 0
' I •,.. , _, ..... , \ \ \ \
I I Classical ' /
., ', , I plate theorv \ ~ I
') ·1 ' ' ... 1 "o ' l 2aM I q b -h '- •• 0 .1 }
y 0 ·~ :
3 4 5 6 7 8
Nond imens ion;1 l ized time, t
(b) Bending moment ve-::-st1s time
Figure 2 The transient resronse <Jf a simply suprorted rectan-gu~'1r isotropic pl:1te subject('d to s:1ddenly ilprlied pat..:11 LL1:1Jing at the cenlcr(Ll,\Ti\ 2, C.:,t~ mL'Sh, /\t=. l)
I
33
Here a and b are the length and width; h is the thickness; p, 'J, E are
the density, Poisson's ratio, and Young's modulus; q 0 1s the uniform
applied load magnitude; and ot is the time-step size. Note that the ra-
tio of b/h is only 10, whereas classical plate theory generally assumes a
ratio of at least 10. Two important results are evident in Figure (9).
First, the classical theory is inadequate for predicting thick-plate re-
sponse. Second, the present program shows excellent agreement with
the analytical thick-plate solution.
Akay (13) obtained both linear and nonlinear results for isotropic
plates using a mixed formulation. In Figure (10), linear results from
this program are compared with Akay·s linear results for a simply sup-
ported square plate subjected to uniform pulse loading. The data for
this case was
a = b = 25 cm, h = 5 cm, 2
" = 8 x 10- 6 Ns -I'-' c;;;1+ ' ') =
i5t = 5, 10 µsec, N qo = cmz '
0.25, E = 2. 1 x 10 ~ cm- (4.2.2)
Although the peak values show good agreement, note that there is a
small phase difference between the two solutions. Overall, agreement is
good. Again, the classical plate theory is presented to illustrate the
significant error which may occur if transverse shear strains are neg-
lected.
As a final verification, a nonlinear analysis was performed and
the results were compared to those reported by Akay for a simply sup-
ported square plate subjected to uniform pulse loading. The data for
this problem was as follows:
a = b = 243.8 c:n' h = 0.635 cm, \) = 0.25, E = 7.031 x 10 5 N Cri]Z"•
2.547 x 10- 5 Ns 4.882 10- 4 N r5 t 0.005 q = w' qo = x cm2 '
< s - (4.2.3)
El u
r' 0
~ . c: 0 .... ._, u
"' .-. ... "' ~:;)
... CJ .... c:: "' u
2 • f),- --T- - -,-- --.--------.
1. 8 <I ~ 10 N/cm 2 0
1. 6
"' 1. 4 E u -z
l. 2 x
D
l. 0 ,; ., 0.8 "' ...
u ., 0.6 ...
"' .., c:: "' u
0.
-0. 2l __ J__ _ _l _ _i__L __ ___!____j_ __ 1
o 80 160 240 120 400 480 s20 ~no
Time, t (li,;.,c.)
(a) Center defll:'ction
200
180
160
140
120
100
80
60
40
LO
0
-20
--r---r-~-.--~---.-~--.r-~-.---~--....
80
q • 10 N/cm 2 0
160 240 320 400 480 520 600
Time, t (~sec.)
(b) Center stress
Fi~~ure 1. Transient response of a simply supported square plate subjected to uniform pulse loading (DATA 1, h:2 mesh, tit = 5JJ sec.)
LEGEND: presL•nt FEM i\t=lO\l mixed FEM (Akay (1980)) i\t 5µ tit=511
w +:>
35
The results are plotted in Figure (lla), along with classical results due
to Volterra and Zachmanoglou. Excellent agreement ~etween the two
nonlinear results are evident, although the present program predicts a
slightly higher peak deflection than Akay. Note that the classical theo-
ry predicts a much larger response and a longer period. This differ-
ence may be attributed to the midplane stretching. The nonlinear re-
sults include the effect of midplane stretching, which increases the
plate's bending stiffness because force is required to stretch the mid-
plane while the plate is bending. The classical theory ignores this ef-
fect, making the plate appear less stiff. The 'stiffer' plate from the
nonlinear analysis has a higher frequency and a lower peak deflection,
as expected.
In Figure (11b), the results for several load magnitudes are com-
pared with Akay. Excellent agreement was obtained. Note that the
peak deflections form a nonlinear curve, clearly exhibiting the effect of
midplane stretching. As the deflection increases, the plate becomes
stiffer in bending because of the extra force required to stretch the
midplane. Using classical linear theory, the peak deflection is a linear
function of the load magnitude.
w( <:m} .9
-~ , ' .l:l.I / \ opresrnt FEM
I l
. I 1 dassical \ (Vol terr a and
.61 1 o 1 Zacttmano9lou
' '\ \ (1965}) .5~ f' I
.4, I . ,.. ~,--.. ~-kay (1980)
Lj \ \ .t·l J ~ . l ~ \ ~
\ J \ .0 k,_. ·-· ov _':, ...... t
Q
00
--·r ·-·-·r·-.----.-
: '~ Lf;· (,;t ~ ,()()25)
l.4
l '2·-
\) r ' 5 ( . t - Qf)"" • r ~/il .. - ... cl}
~ :' . . ~s~q i M • __ .006)
J ·"] ' -~ .... y'l ( ·.t f I / ': \
1~ / / . )~ ,·
£(~;.~,,, !i .. ~.C.-4 • .1-.
l. ·1 .3 .
.6 . '
4
. ? -
0.
. 00';)
....... +·· . 05 . I . 15 . 2 . 25
Tim~, t . 12 ";· '. 1-4, . l 6
(a)
(b)
Figure 10 Comparison with Non1 inear Results from AKay.
Nonlinear transient response of a simply supported
square plate under a uni fonnly disti rbuted load.
Data is from eqn. 4.2.3.
\.<) a'>
37
'1.3 EFFECT OF ROTATORY INERTIA AND TRANSVERSE SHEAR CONSTANT
In this section, the effect of different transverse shear constants
and the effect of including rotatory inertia are illustrated. It is impor-
tant to recall from the previous discussion that these effects are highly
dependent on the particular problem for which they are investigated.
Specifically, in the case of transverse shear constant the effects are
more pronounced when the transverse shear constant is small compared
to the in-plane Young's modulii. Rotatory inertia effects are more pro-
nounced in thick plates or in large deflection analysis. Either effect
may or may not be significant in any particular problem. It is this un-
certainty which makes including these effects so important in analysis
work.
In Figure (11), the effects of the transverse shear constant are
illustrated for a specific material and a specific load. The material
properties for this case correspond to a Graphite-Epoxy plate with a
64% fiber volume fraction. The data for this comparison is as follows:
E1 = 22.2 msi, E2 = 1.54 msi, a= b = 1 in, h=0.1 in
Two-1 ayer, cross-ply q = ~ E2h = 15 (4.3.1)
CASE 1 : G23 = G12 = G13 = 0.98 msi
CASE 3: G12 = G13 = 0.98 msi, G23 = 0.54 msi
Note that the difference between the two predicted responses in Figure
(11) is obvious, but not too large. Again, it should be reiterated that
z 0
1-u
"'tW ~_J ...
38
EFFECT OF TRANSVERSE SHEAR CONSTANT - TWO CASES CASE [ll: G23 = Gl2, CASE (2l: G23 = 1/2 G12
TWO-LAIER CROSS-PL I, UOL, SS, SQ. PLATE WITH A/H=lO
=r I ('l")i-•
1 = - G12 2 -
LL. w C\J
0
-
I I 1 2 3 LJ
TIME 1110·!
Figure 11 Effect of Transverse Shear Constant--A Specific Example.
Data for these cases is presented in eqn. (4.3.1), and
corresponds to Graphite-Epoxy with a 64% fiber volume
fraction. Note that the effect may be greater or less
for other data cases.
39
the difference may be greater or smaller for another data case, and it
is precisely this uncertainty which necessitates including the transverse
shear deformation in the analysis.
The effect of supressing the rotatory inertia term (i.e., artificial-
ly forcing I to zero in the equations of motion) is illustrated in Figure
(12). The data for this case is given below:
E1 = 25 x 10 6 , Ez = x 10 6 p = 1 ' 'J = 0.25 ' G12 = G 13 = G12 = 0.5 x 10 6 a = b = 1 ' h = 0. 1 '
~ Two-1 ayer, cross-ply, q = Eza = 100 (4.3.2)
Simply supported, uniformly distributed load
For this data case, the effect of rotatory inertia is quite pronounced.
With the rotatory inertia suppressed, the maximum predicted deflection
is much smaller because the effect of the inertia of rotation is ignored.
Only the transverse inertia is accounted for. In small deflection analy-
sis, the rotations are small enough that often the rotatory inertia may
be ignored without significantly affecting the predicted response. How-
ever, if the deflections are large, the inertia of the rotations may have
a greater effect on the solution, as illustrated in Figure (12). The ex-
act contribution of the rotatory inertia will be different for various
problems. Again, it is this uncertainty which requires rotatory inertia
to be included in the analysis.
The effect of including the geometric nonlinearities 1s amply illus-
trated in the previous section and is not repeated here.
40
EFFECT OF ROTARY I~ERTIA - I SET TO ZERO IN ONE CRSE T W 0 - L R I E 8 C Fl 0 S S P L '( , U D 1_ • S ~' , S CL P L P. T E ~: IT :-i h I :-1 = 1 0
QBRR=lGO, E1/E2=25
~'- 11 _,
0
z I 0 ,,~ --' ::i'}-~ I l I- included '}' u I
~ ~w I .. I _J LL. w (\I._
0 !
Figure 12 Effect of Rotary Inertia in a Specific Example.
The data for these cases is given in eqn. (4.3.2).
41
There is an interesting interaction between the effects of the
transverse shear constant and the rotatory inertia. The rotary inertia
is much more pronounced in large deflection analysis and tends to mask
the effect of transverse shear deformation in large-deflection problems.
On the other hand, the rotatory inertia may be negligible for small-de-
flection analysis whereas the transverse shear deformation effect be-
comes much more pronounced. Thus, the effect of one of these cons id-
erations may be hidden by the effect of the other in any given
problem. Note that in Figure (11), the normalized load magnitude q is
15, but in Figure (12) the q is 100. A run using the data for Figure
(11) with the rotatory inertia suppressed produced a negligible differ-
ence from the curve in Figure (11); a run with the different transverse
shear constants using q of 100 gave identical results for the two con-
stants. It is essential to include both effects since the outcome for a
specific case cannot be predicted in advance.
'I.LI PARAMETRIC RESULTS
With the program's accuracy verified in the previous section, a
study of the effects of various parameters is presented in this section.
The mesh shown in Figure (Sa) was used in all of these cases, along
For a cross-ply laminate (0 °/90°), Oi 6 and 0 26 are zero, effectively
uncoupling the shear quantities from the normal quantities. Thus, the
plate exhibits less bending stiffness and shows a greater deflection for
a given load magnitude (see Figure 16). Although eqn. (4.3. 1) strictly
applies only to plane stress within a layer, the uncoupling phenomenon
occurs in the general case.
In Figure ( 17), the effect of plate thickness is plotted. Note
that the thin plate has a much higher peak deflection than the thick
plate for the same normalized load magnitude q. This is another indica-
tion of the effect of transverse shear and geometric nonlinearities,
which are much more pronounced in thick plates than in thin. The ef-
fect of including geometric nonlinearities is to increase the bending re-
sistance in thick-plate or large defelection analysis. Figure (17) has
47
EFFECT OF LAMINATION SCHEME - CP, AP=5,15,25,35,45 TWO-LAYER CROSS-PL I, UDL, SS, SQ. PLATE WITH A/H=lO
QBRR=SO, El/E2=25
I/)_
z ::I'
0 t--4
I-u Ct'>
~w :! _J • LL w (\J
0
-
1 2 3 TIME
110·3
Figure 16 Effect of Lamination Scheme.
The A.P. plates are ±5°, ±15°, ±25°, ±35°,
±45°. Al 1 A.P. plates showed greater
greater stiffness than the C.P. plate. This
may be attributed to the Qis, Q15 terms as
discussed on page 46.
48
the normalized deflection plotted for identical normalized loads, and
clearly the thick plate exhibits much greater bending stiffness than the
thin plate. The thick plate has a higher frequency and a lower peak
deflection.
The effects of boundary conditions, Newmark parameters ,J. and ;3,
and time-step size were previously discussed in Chapter 111 and will not
be repeated here.
49
EFFECT ~F PLATE THICKNESS - (A,8l /H = 10, 20 T W 0 - L A I C: R C R 0 S S - P L '( , U 0 L , S S , S Q • P L A T E vll T H A I H = 1 0
OBAR=50, E1/E2=25
~~1 g~, = 20
fJ_ (I")!_ "i'W ~o
JOI
0 (\J
,.. _J ([ :E: T-a: D 10
z 2
ME xl 0·3
Figure 17 Effect of Plate Thickness.
The normalized deflection is much less for
the thick plate because of nonlinear effects,
transverse shear deformation, and rotatory
inertia. The thick plate exhibits much
greater stiffness than the thin plate.
Chapter V
SUMMARY AND CONCLUSIONS
5.1 SUMMARY
In the introductory chapter, the effects of transverse shear de-
formation and geometric nonlinearities were shown to be potentially sig-
nificant in the bending response of laminated anisotropic plates. In the
results presented throughout the thesis, this significance was proven
for a number of specific cases. These effcts are especially pronounced
for large deflections, thick plates, or plates with high ratios of the
in-plane Young's modulus to the transverse shear modulus.
Classical theory has proven to be inadequate in predicting the
bending response of such plates because the theory ignores transverse
shear deformation and geometric nonlinearity. Numerous alternate theo-
ries have been proposed suggesting several different methods of ac-
counting for these effects. In this thesis, a theory due to Yang, Nor-
ris, and Stavsky was used to account for transverse shear deformation
and rotatory inertia. The von Karman assumptions were used to ac-
count for geometric nonlinearities, and Newmark's direct integration
technique was employed to integrate the resulting equations of motion.
Next, a computer program was developed which incorporates the
theory and approximations as outlined in the text. The program is
simple to use and has general applications for analysis work. Other
50
51
analysts could use the program in its present form to solve a range of
problems. Such capabilities as accounting for material nonlinearities can
be added (in this case, to the STIFF subroutine) without much difficul-
ty, and without disturbing the execution logic.
The program was used to analyze the effects of load magnitude,
material orthotropy, plate aspect ratio, plate thickness, and lamination
scheme on the bending response of two-layer unsymmetric laminated
plates.
5. 2 CONCLU SIGNS
The method developed in this thesis (also reported in <10>) gives
excellent results in the transient bending analysis of laminated compos-
ite plates. The computer program is accurate, efficient, and general,
and apparently represents the first transient finite-element analysis of
laminated composite plate bending which accounts for transverse shear,
rotatory inertia, and geometric nonlinearities. These effects may give
solutions which differ significantly from the classical solutions, as
shown in Chapter IV. Thus, this program will be valuable in analysis
work. The benefits are especially useful when analyzing thick-plate re-
sponse, large deflections, or materials with a high ratio of the in-plane
Young's modulus to the transverse shear modulus. Many problems of
practical importance fall into one or more of these categories.
Although the program 1s currently written for transversely iso-
tropic materials with constant material properties, it may be easily
52
adapted to more anisotropic materials or to materials with material nonli-
nearities. This 1s accomplished by modifying the stiffness subroutine.
The basic program logic which assembles and solves the global equations
1s not affected by the type of element used.
Finally, to reiterate, the program developed in this thesis works
well and represents the first transient finite-element analysis of laminat-
ed composite plate bending which accounts for transverse shear defor-
mation, rotatory inertia, and geometric nonlinearities.
REFERENCES
1. Yang, P. C., Norris, C. H., and Stavsky, Y. 'Elastic Wave Propagation in Heterogeneous Plates', International Journal of Solids and Structures, Vol. 2, 1966, pp. 665-684.
2. Reissner, E. 'The Effect of Transverse Shear Deformation on the Bending of Elastic Plates,' Journal of Applied Mechanics, Vol. 12, No. 2, Trans. ASME, Vol. 67, June 1945, pp. 69-77.
3. Mindlin, R. D. 'Influence of Rotatory Inertia and Shear on Flexural Motions of Isotropic, Elastic Plates,· Journal of Applied Mechanics, Vol. 18, No. 1, Trans. ASME, Vol. 73, Mar. 1951, pp.31-38
4. Stavsky, Y. 'On the Theory of Symmetrically Heterogeneous Plates Having the Same Thickness Variation of the Elastic Moduli', Topics in Applied Mechanics, Ed. by Abi r, D., Ollendorff, F. and Reiner, M., American Elsevier, New York, 1965.
5. Ambartsumyan, S. A. Theory of Anisotropic Plates translated from Russion by Cheron, T., and ed. Ashton, J. E., Technomic, 1969.
6. Pryor, Jr., C. W., and Barker, R. M. 'A Finite Element Analysis Including Transverse Shear Effects for Applications to Laminated Plates', A/AA Journal, Vol. 9, pp. 912-917, 1971.
7. Noor, A. K. and Hartley, S. J. 'Effect of Shear Deformation and Anisotropy on the Non-Linear Response of Composite Plates', Developments in Composite Materials - 1, Ed. Holister, G., Applied Science Publishers, Barking, Essex, England, pp. 55-65, 1977.
8. Reddy, J. N. 'A Penalty Plate-Bending Element for the Analysis of Laminated Anisotropic Composite Plates', Int. J. Numer. Meth. Engng., Vol. 15, pp. 1187-1206, 1980.
9. Reddy, J. N. and Chao, W. C. 'Nonlinear Bending of Thick Rectangular, Laminated Composite Plates', Int. J. Nonlinear Mechanics, 1981, to appear.
10. Reddy, J. N. and Mook, D. J. 'Transient Analysis of Layered Composite Plates Using A Shear Deformation Theory', to be presented at the Int' I. Conf. on Computational Methods and Experimental Measurements, Washington, D. C., June 30-July 2, 1982.
11. Volterra, E. and Zachmanoglou, E. C. Dynamics of Vibrations, Merrill, Columbus, Ohio, 1965.
53
54
12. Reismann, H. and Lee, Y. 'Forced Motions of Rectangular Plates', Developments in Theoretical and Applied Mechanics, Vol. 4, D. Frederick (ed.), pp. 3-18, Pergamon Press, New York, 1969.
13 · Akay, H. U. 'Dynamic Large Deflection Analysis of Plates Using Mixed Finite Elements', Computers and Structures, Vol. 11, pp. 1-11, Pergamon Press, 1980.
14. Jones, Robert M. Mechanics of Composite Materials, Scripta Book Company (McGraw-Hill), Washington, D.C., 1975.
15. Ashton, J. E., and Whitney, J. M. Theory Of Laminated Plates, TECHNOMIC Publishing Co., Stamford, CT, 1970.
16. Tsai, Steven W. Mechanics of Composite Materials, Part II, Theoretical Aspects, Air Force Materials Laboratory Tech. Report AFML-TR-66-149, November, 1966.
17. Whitney, J. M. and Pagano, N. J. 'Shear Deformation in Heterogeneous Anisotropic Plates', J. Applied Mechanics, Vol. 37, pp. 1031-1036, 1970.
18. Bathe, K. J. and Wilson, E. L. Numerical Methods In Finite Element Analysis, Prentice-Hall, Englewood Cliffs, New Jersey, 1976.
[Ks s J = D2s([sxy] + [sxy]T) + Ds&[Sxx] + D22[sYYJ + Ai+i+[S]. and
58
J: acp. 3¢. s~~ s .. = a;' ~ dxdy, s,1 = O,x,y, - s .. '
I J Re oil IJ I j
R:n =he t (~~) 3¢. d<jl. I~
~ '~ 'T1 = O,x,y, ~ , dxdy, ~ a~ a :J
A11(~:) 2
A12(~w) 2 aw cw N1 = + + 2A16 dX ay• oy
Aid~:) 2
A2d~~) 2 aw dW N2 = + + 2A25 dX oy'
A (dW) 2
A26(~~) 2 ow 3w N5 = 16 3x + + 2As6 dX ay
Appendix 8
FLOW DIAGRAM OF COMPUTER PROGRAM
59
60
I INPUT DATA l
1 I CALCULATE MATERIAL CONSTANTS I
1 I GENERATE MESH k----
1 I CALCULATE GEOMETRY CONSTANTS I
1 CALCULATE CONSTITUTIVE
MATRICES [QJ) [AL [BL CDJ
1 CALCULATE TIME CONSTANTS
CNEWMARK COEFFICIENTS)
1 I BEGIN TIME LOOP l
61
...--..---->!BEGIN TIME STEP I
l ~~> CALCULATE ELEMENT MATRICES
I ASSEMBLE [KI J J {FI} I
l I IMPOSE THE BI c I Is I
l I SOLVE FOR THE DISPLACEMENTS ~ . l
l FA! LS I I CRITERION CHECK CONVERGENCE I
MEETS CRITERION
+ I CALCULATE VELOCITY AND ACCELERATION I
l CALCULATE STRESSES > AND BENDING MOMENTS
l k---"I RETURN FOR NEXT TIME STEP IF APPLICABLE I
l ~
Appendix C
LIST/NC OF COMPUTER PROGRAM
62
63
C **********************************************************************MAl00010 C MAl00020 C PROGRAM FOR TRANSIENT ANALYSIS OF LAMINATED COMPOSITE PLATES MAI00030 C ACCOUNTING FOR TRANSVERSE SHEAR STRAINS AND GEOMETRIC NONLINEARITIESMAI00040 C MAI00050 C ORIGINAL VERSION DEVELOPED BY J.N. REDDY MAI00060 C MAl00070 C REVISED BY D.J. MOOK MAI00080 C MAI00090 C MAI00100 C QUANTITIES TO BE INPUT ARE AS FOLLOWS: MA100110 C MAf 00120 C CARD NO. FORMAT VARIABLES MAI00130 C MAl00140 C 615 LAYER; NO. OF LAYERS MAI00150 C NLS; NO. OF CASES (LOAD STEPS) TO BE RUN MAl00160 C INTER; SET EQUAL TO 1 MAI00170 C NPRNT; =1, CAUSES ELEMENT MATRICES TO BE PRINTEDMAI00180 C !LOAD; =O FOR U.D.L MAl00190 C =1 FOR SINUSOIDAL QUARTER-PLATE LOADING MAl00200 C ITMAX; LIMIT ON NUMBER OF ITERATIONS AT ANY TIMEMAI00210 C MAI00220 C 2 515 IEL; =1, LINEAR ELEMENT MAl00230 C =2, QUADRATIC ELEMENT MAI00240 C NOF; NUMBER OF DEGREES OF FREEDOM PER NOOE MAf 00250 C NPE; NODES PER ELEMENT MAI00260 C NRMAX; MAXIMUM ROW DIMENSION (=TOTAL D.O.F.) MAI00270 C NCMAX; MAXIMUM COLUMN DIMENSION (= BANDWIDTH) MAI00280 C MAl00290 C 3 1615 NTIME; INPUT 'NLS' TIMES - NUMBER OF TIME STEPS MAl00300 C MAl00310 C 4 215 !MESH; =1 FOR INPUT OF MESH INFORMATION; MAI00320 C OTHERWISE, AUTOMATIC MESH GENERATION MAI00330 C IVAL; =1, INPUT NONZERO BOUNDARY CONDITIONS MAI00340 C MAf 00350 C 5 8F10 THETA; INPUT 'LAYER' TIMES - ANGLE OF EACH LAYERMAI00360 C MAI00370 C 6 8F10 H; TOTAL THICKNESS OF PLATE MAI00380 C DH; INPUT 'LAYER' TIMES - THICKNESS OF LAYER MAI00390 C MAI00400 C THE NEXT CARO IS INPUT FOR AUTOMATIC MESH GENERATION ( IMESH NOT 1) MAI00410 C MAI00420 C (7) 215,2F10 NX; NUMBER OF ELEMENTS DESIRED IN X-DIRECTION MAI00430 C NY; NUMBER OF ELEMENTS DESIRED IN Y-OIRECTION MAl00440 C XL; LENGTH OF RECTANGLE IN X-DIRECTION MAI00450 C YL; LENGTH OF RECTANGLE IN Y-OIRECTION MAI00460 C MAl00470 C THE NEXT THREE (3) CARDS ARE INPUT IF 'IMESH' IS 1 MAI00480 C MAl00490 C (7) 215 NNODE; NUMBER OF NODES I~ THE MESH MAI00500 C NEM; NUMBER OF ELEMENTS IN THE MESH MAI00510 C MAI00520
64
C (8) 2F10 X; INPUT 'NNODE' TIMES - X-COORDINATES OF NODES MAI00530 C Y; INPUT 'NNODE' TIMES - Y-COORDINATES Of NODES MA100540 C MAI00550 C (9) 1615 NOD( l,J); CONNECTIVITY MATRIX. NOD( 1,J) MAI00560 C CONTAINS THE GLOBAL NUMBER OF NODE J MAl00570 C IN ELEMENT I. MAI00580 C MAI00590 C 10 7Fl0 RHO; DENSITY OF THE MATERIAL MAI00600 C El,E2; YOUNG'S MODULI I IN MATERIAL COORDINATES MAI00610 C G12,G13,G23; SHEAR MODULI I IN MATERIAL COORDS. MAI00620 C ANU12; POISSON'S RATIO IN MATERIAL COORDINATES MAI00630 C MAI 00640 C 11 8F10 DTM; INPUT 'NLS' TIMES - TIME-STEP SIZE MAl00650 C MAl00660 C 12 8Fl0 PO; LOAD MAGNITUDE FOR INITIAL CASE MAl00670 C DP; INPUT 'NLS-1' TIMES - LOAD INCREMENT MAI00680 C MA I 00690 C 13 lFlO EPS; MAXIMUM ALLOWABLE ERROR FOR NONLINEAR ITER.MAI00700 C MAl00710 C 14 3F10 ALFA; NEWMARK-BETA CONSTANTALPHA MAl00720 C GAMA; MAl00730 C AK; SHEAR CORRECTION COEFFICIENT MAl00740 C MA I 00750 C 15 15 NBDY; NUMBER OF BOUNDARY CONDITIONS MAI00760 C MA100770 C THE NEXT CARD IS INPUT ONLY If NBDY IS GREATER THAN ZERO MAI00780 C MAI00790 C (16) 1615 IBDY; INPUT 'NBDY' TIMES - IDENTIFIES NODAL B.C.MAl00800 C MAl00810 C THE NEXT CARD IS INPUT ONLY IF !VAL IS MAI00820 C MAI00830 C (17) 8Fl0 VBDY; INPUT 'NBDY' TIMES - VALUE Of B.C. IBDY MAl00840 C MAl00850
* Gf2(125),DP(l1),DTM(5),NTIME(5) MA 00890 DI MENS I ON VBDY ( 4 7) , I BOY ( 4 7) , A ( 6, 6 ) , C ( 3, 3 ) , TH ET A ( 16 ) , DH ( 5 ) MA 00900 COMMON/STIF1/ELXY(9,2),STIF(45,45),ELP(45),W(45),W0(45),W1(45), MA 00910
* W2(45),AO,Al,A2,A3,A4,C1,C2,XL,YL MA 00920 COMMON /MAT/Q( 16,3,3),Z(17),Q44(16),Q45(16),Q55(16) MA 00930 COMMON /MES1/X(81),Y(81),NOD(16,9) MA 00940 READ(5,1000) LAYER,NLS, INTER,NPRNT, ILOAD, ITMAX MA 00950 READ(5,l000) IEL,NDF,NPE,NRMAX,NCMAX MA 00960 READ(5,1000) (NTIME( I), l=l,NLS) MA 00970 READ(5, 1000) IMESH, IVAL MA 00980 READ(5,1001) (THETA(l),l=l,LAYER) MA 00990 READ(5,1001) H,(DH(l),1=1,LAYER) MAOlOOO IF( !MESH. EQ. l) GO TO 1 MA 01010 READ(5,1002) NX,NY,XL,YL MAl01020 GO TO 2 MAl01030
3 YBOY( I )=O. DO IF(IYAL.EQ.1) READ(5,1001) (VBOY(l),1=1,NBDY)
4NGP=IEL+1 LGP = I El BETA=0.25*(0.5+ALFA)**2 ANU21=ANU12*E2/E1 ALMDA=1.0-ANU12*ANU21
MA 01080 MA 01090 MA 01100 MA 01110 MA 01120 MA 01130 MA 01140 MA 01150 MA 01160 MA 01170 MA 01180 MA 01190 MA 01200 MA 01210 MA 01220
C MA 01230 C **********************************************************************MA 01240 C MA 01250 C CALCULATE THE STIFFNESS MATRIX IN MATERIAL PRINCIPLE COORDINATES MA 01260 C MA 01270 C **********************************************************************MA 01280 C MA 01290
CALL SUBROUTINE MESH FOR MESH GENERATION MAl01440 MA101450 MAl01460
C **********************************************************************MAI01470 C MAI01480
DO 30 1=1,NBOY MAI01490 30 YBOY( I )=0.0 MA 01500
NEQ=NNM*NDF MA 01510 WRITE(6,420) NEM,NNM,NOF MA 01520 NN=NPE*NDF MA 01530 WRITE(6,360) NBDY MA 01540 WRITE(6,270) ( IBOY( I), 1=1,NBDY) MA 01550 WRITE(6,290) MA 01560 DO 40 1=1,NEM MA 01570
40 WRITE(6,270) l,(NOD( l,J),J==1,NPE) MA 01580 WRITE(6,350) MA 01590 WRITE(6,280) (X( I ),Y( I), 1=1,NNM) MA 01600
C MA 01610 C **********************************************************************MA 01620
c c c
66
CALCULATE THE BANDWIDTH, HALF-SW, NO. OF EQNS., ETC MAI01630 MAI01640 MAl01650
C **********************************************************************MAl01660 C MAI01670
NHBW=O DO 50 N=l,NEM DO 50 1=1,NPE DO 50 J=l,NPE NW=( IABS(NOD(N, I )-NOO(N,J))+l)*NOF
50 IF (NHBW.LT.NW) NHBW=NW WRITE(6,430) NHBW NBW=2*NHBW
MAl01680 MAl01690 MAI01700 MA 01710 MA 01720 MA 01730 MA 01740 MA 01750 MA 01760 DO 240 NH=l,1
C MA 01770 C **********************************************************************MA 01780 C MA 01790 C CALCULATE PANO I (ROTATORY INERTIA) MA 01800 C MA 01810 C **********************************************************************MA 01820 C MAI01830
C **********************************************************************MAl01890 C MAI01900 C CALL SUBROUTINE MATPRP TO CALCULATE QBAR MATRIX, AND A,B,D MATRICES MAl01910 C MA I 01920 c c
CALL MATPRP {LAYER,C,A,H,THETA,AK,Gl3,G23,A44,A45,A55) WR I TE ( 6 , 3 90 ) DO 230 NP=l,NLS PBAR=(2.0*XL)**4*PO/E2/(H**4) WRITE(6,340) NP,OP(NP),PO,PBAR OT=DTM(NP) DT2=DT*OT AO=l.O/BETA/DT2 A2=1.0/BETA/OT Al=ALFA*A2 A3=0.5/BETA-l.O A4=ALFA/BETA-l.O WRITE(6,540) ALFA,BETA,AO,Al,A2,A3,A4 WR I TE ( 6, 5 71 )
MA101940 MAl01950 MAI01960 MAI01970 MA101980 MAI01990 MAI02000 MAI02010 MAl02020 MAl02030 MAl02040 MAl02050 MAI02060 MA I 02070 MAI02080
C MA I 02090 C **********************************************************************MAl02100 C MAI02110 C SET THE INITIAL CONDITIONS MA102120 C MAl02130 c c
70 ITER=ITER+1 IF ( ITER.GT. ITMAX) GO TO 185 DO 80 1=1,NEQ GP( I )=GF( I) GF( I )=GSTI F( I, NBW) DO 80 J=1,NBW
MAI02330 MAl02340 MAI02350 MA 02360 MA 02370 MA 02380 MA 02390 MA 02400 MA 02410 MA 02420 MA 02430 80 GSTIF( l,J)=O.O
C MA 02440 C **********************************************************************MA 02450 C MA 02460 C BEGIN STIFFNESS MATRIX ITERATION FOR THIS TIME MA 02470 C MA 02480 C **********************************************************************MA 02490 C MA 02500
DO 1 3 0 N= 1 , N EM L=O DO 90 1=1,NPE NI =NOD( N, I ) ELXY( I, 1 )=X( NI ) ELXY( I ,2)=Y(NI) Ll=(Nl-1)*NOF DO 90 J=1,NOF Ll=Ll+1 L=L+l WO( L)=GFO( LI) Wl ( L)=GF1 (LI) W2(l)=GF2(LI) W(L)=GAMA*GP(LI )+(1.0-GAMA)*GF(LI)
MA 02510 MA 02520 MA 02530 MA 02540 MA 02550 MA 02560 MA 02570 MA 02580 MA 02590 MA 02600 MA 02610 MA 02620 MA 02630 MA 02640 MA 02650 90 IF(ITMAX .EQ. 1)W(L)=O.O
C MA 02660 C **********************************************************************MA 02670 C MA 02680 C CALL SUBROUTINE STIFF TO CALCULATE THE ELEMENT MASS ANO STIFFNESS MA 02690 C ANO STIFFNESS MATRICES ANO THE ELEMENT FORCE VECTOR MA 02700 C MA 02710 C THE MODIFICATIONS FOR THE NEWMARK DIRECT INTEGRATION TECHNIQUE MA 02720
c c
68
ARE CARRIED OUT AT THE ELEMENT LEVEL IN SUBROUTINE STIFF MAI02730 MAI02740
C **********************************************************************MAI02750 C MA 02760
100
110
CALL STIFF(NPE,NN,NGP,LGP, ILOAD,NT,A,A44,A45,A55,PO, ITER,N) IF (NPRNT.EQ.O) GO TO 110 IF (N.GT.1) GO TO 110 WRITE(6,320) DO 100 1=1,NN WRITE(6,300) (STIF( l,J),J=l,NN) WRITE(6,310) WRITE(6,300) (ELP( I), 1=1,NN) WRITE(6,310) CONTINUE
MA 02770 MA 02780 MA 02790 MA 02800 MA 02810 MA 02820 MA 02830 MA 02840 MA 02850 MA 02860
C MA 02870 C **********************************************************************MA 02880 C MA 02890 C ASSEMBLE THE ELEMENT STIFFNESS MATRICES TO GET THE GLOBAL MA 02900 C STIFFNESS MATRIX ANO THE ELEMENT FORCE VECTORS TO GET THE GLOBAL MA 02910 C FORCE VECTOR. MAI02920 C MA I 02930 C **********************************************************************MAI02940 C MAI 02950
120 130
DO 130 1=1,NPE NR=(NOO(N, I )-l)*NDF DO 130 11=1,NDF NR=NR+l L=( 1-1 )*NDF+l I GSTIF(NR,NBW)=GSTIF(NR,NBW)+ELP(L) DO 130 J=l,NPE NCL=(NOO(N,J)-1 )*NDF DO 130 JJ=l,NOF M=(J-l)*NDF+JJ NC=NCL+JJ-NR+NHBW IF (NC) 130,130,120 GSTIF(NR,NC)=GSTIF(NR,NC)+STIF(L,M) CONTINUE
C MA103100 C **********************************************************************MA103110 C MAl03120 C IMPOSE THE BOUNDARY CONDITIONS MA 03130 C MA 03140 C **********************************************************************MA 03150 C MA 03160
140
150 c
Nl=NDF-1 N2=NDF-2 I F ( I LOAD. EQ. 0 ) 00 150 1=1,NBDY I B= I BOY( I) VB=VBDY( I) DO 140 J=l,NBW
GSTIF(N2,NBW)=GSTIF(N2,NBW)+0.25*PO
GST I F ( I B, J ) =O. 0 GST I F ( I B, NHBW) = 1 . 0 GSTIF( IB,NBW)=VB
MA 03170 MA 03180 MA 03190 MA 03200 MA 03210 MA 03220 MA 03230 MA 03240 MA 03250 MA 03260 MA 03270
69
C **********************************************************************MAI03280 C MAI 03290 C CALL SUBROUTINE BOUNSM TO SOLVE THE SYSTEM OF SIMULTANEOUS MAI03300 C EQUATIONS. THE SOLUTION (THE NOOE-POINT DISPLACEMENTS) IS MAl03310 C RETURNED IN THE LAST COLUMN OF THE COEFFICIENT MATRIX (THE MAI03320 C ASSEMBLED, MODIFIED GLOBAL STIFFNESS MATRIX) MAl03330 C MAI03340 c c
CALL BOUNSM (GSTIF,NRMAX,NCMAX,NEQ,NHBW) ERR=O.O IF ( ITMAX.EQ.1) GO TO 180 00 170 l=N2,NEQ,NOF ERR=(GSTIF( 1,NBW)-GF( I ))**2+ERR ERR=DSQRT(ERR)/OABS(GSTIF(N2,NBW)) IF (ERR.GT.EPS) GO TO 70 GOTO 185 WRITE(6,550) ITER CONTINUE
C **********************************************************************MAI03480 C MAI03490 C CALCULATE THE VELOCITY AND ACCELERATION VECTORS USING THE DIS- MAI03500 C PLACEMENT VECTOR ANO EQNS. 3.4.1 FROM SECTION 3.4 MAI03510 C MA I 03520 C **********************************************************************MA 03530 C MA 03540
DO 65 1=1,NEQ GFO( I )=AO*( GST IF( I, NBW)-GFO( I) )-A2*GF1 (I )-A3*GF2( I) GF1 (I )=GF1 (I )+OT*( 1.0-ALFA)*GF2( I )+OT*ALFA*GFO( I) GF2( I )=CFO( I) CFO( I )=CSTI F( I ,NBW)
65 CONTINUE
190
IF(NPRNT .EO. O)GOTO 195 WRITE(6,560) WRITE(6,530) NT,DT,T IF (NOF.EQ.3) GO TO 190 WRITE(6,380) WR I TE ( 6, 300) (CST I F ( I , NBW), I= 1, NEQ, NOF) WR TE(6,310) WR TE(6,300)(GF1( I), 1=1,NEQ,NDF) WR TE(6,310) WR TE(6,300)(GF2( I), 1=1,NEQ,NDF) WR TE(6,440) WR TE(6,300)(GSTIF( l,NBW), 1=2,NEQ,NOF) WR TE(6,310) WR TE(6,300)(GF1( I), 1=2,NEQ,NOF) WR TE(6,310) WR TE(6,300)(GF2( I), 1=2,NEQ,NDF) WR TE(6,450) WR TE(6,300)(GSTIF( l,NBW), l=N2,NEQ,NDF) WR TE(6,310) WR TE(6,300)(GF1( I), l=N2,NEQ,NOF) WR TE(6,310) WR TE(6,300)(GF2( I), l=N2,NEQ,NDF)
MA 03550 MA 03560 MA 03570 MA 03580 MA 03590 MA 03600 MA 03610 MA 03620 MA 03630 MA 03640 MA 03650 MA 03660 MA 03670 MAI03680 MAl03690 MAI03700 MA103710 MAl03720 MAI03730 MAI03740 MAl03750 MAl03760 MAI03770 MAI03780 MAI03790 MAI03800 MAl03810 MAl03820
MAI03830 MAl03840 MA103850 MAI03860 MA 03870 MA 03880 MA 03890 MA 03900 MA 03910 MA 03920 MA 03930 MA 03940 MA 03950
C MA 03960 IF ( ITMAX.EQ.1) GO TO 200 195
C **********************************************************************MA 03970 C MA 03980 C IF THE SOLUTION AT THIS TIME STEP HAS NOT CONVERGED IN THE MA 03990 C NUMBER OF ITERATIONS SPECIFIED IN THE INPUT, PROGRAM STOPS MA 04000 C MA 04010 C **********************************************************************MA 04020 C MA 04030
200
210 c
IF ( ITER.GT. ITMAX) GO TO 250 IF ( ITER.EQ.1) GO TO 70 WRITE(6,310) CONTINUE DO 220 N=l, NEM L=O DO 21 0 I = 1 , NP E Nl=NOD(N, I) ELXY( I, 1 )=X( NI) ELXY( I, 2 )=Y( NI ) Ll=(Nl-1 )*NDF DO 210 J=1,NDF Ll=Ll+l L=L+l W(L)=GSTIF(Ll,NBW)
MA 04040 MA 04050 MA 04060 MA 04070 MA 04080 MA 04090 MA 04100 MA 04110 MA 04120 MA 04130 MA 04140 MAf 04150 MAl04160 MAI04170 MAI04180 MAI04190
C **********************************************************************MAI04200 c MAl04210 C CALL SUBROUTINE STRESS TO CALCULATE THE 'SECONDARY' QUANTITIES MAI04220 C SUCH AS STRESSES AND BENDING MOMENTS MAl04230 C MAl04240 C **********************************************************************MAI04250 C MAI04260
CALL STRESS (NPE,NDF, IEL,ELXY,W,LAYER, ITER,NP,STRSX,STRXY) MAI04270 I F ( N . EQ. 1 ) STRES=STRSX MA I 04280
220 CONTINUE MAI04290 WRITE(6,570) T,GSTIF(N2,NBW),GF1(N2),GF2(N2),STRES,STRXY,ERR, ITER MAl04300 PT(NT)=T MAI04310 DFL( NT )=GF( N2) MA I 04320 PSTRSX(NT)=STRSX MAl04330 PSTRXY(NT)=STRXY MAI04340 DDFL=OFL(NT) MAI04350
C MAl04360 C **********************************************************************MAl04370
c c c
71
THE SOLUTION IS WRITTEN TO A TAPE FOR LATER USE (SUCH AS PLOTTING) MAI04380 MAl04390 MAl04400
C **********************************************************************MAl04410 C MAI 04420
270 FORMAT (1615) MAI04530 280 FORMAT (8F10.4) MA104540 290 FORMAT (/,4X,' BOOLEAN (CONNECTIVITY) MATRIX-NOD( l,J) ',/) MAI04550 300 FORMAT (8(2X,E12.5)) MAI04560 310 FORMAT (II/ MAI04570 320 FORMAT (/, ELEMENT STIFFNESS MATRICES .... ',/) MA104580 330 FORMAT (/,5X, '*****CONVERGENCE CRITERIA IS NOT SATISFIED*****' ,/MAI04590
380 FORMAT (5X,' .. U-DISPLACEMENT VECTOR.,.',/) MAl04710 390 FORMAT (120('.'),//) MAI04720 400 FORMAT (//,5X, 1 • , .SHEAR SLOPE, Sl-X ... 1 ,/) MAl04730 410 FORMAT (//,5X,' ... SHEAR SLOPE, Sl-Y ... ',/) MAl04740 420 FORMAT (10X, 'ACTUAL NUMBER OF ELEMENTS IN THE MESH=', 13,/,lOX, 'NUMAI04750
1MBER OF NODES IN THE MESH=', 13,/, 10X, 'DEGREES OF FREEDOM=', 12,/)MAl04760 430 FORMAT (//,' HALF BAND WIDTH OF GLOBAL STIFFNESS MATRIX= ', 15,/) MA 04770 440 FORMAT (5X, ' .. V-DISPLACEMENT VECTOR ... ',/) MA 04780 450 FORMAT (5X,' .. W-OISPLACEMENT VECTOR ... ',/) MA 04790 460 FORMAT (/,5X, 'ELEMENT TYPE(1=LINEAR,2=QUADRATIC) =', 12,lOX, 'NGP=' ,MA 04800
11X,12,3X, 1 LGP=',1X,12,/) MA 04810 470 FORMAT (///,2X,'ITERATION NO =',12,5X,'ERROR =',E13.6,//) MA 04820 480 FORMAT (5X, 'THICKNESS=' ,E10.3,2X, 'NUMBER OF LAYERS AND THEIR ORI EMA 04830
lNTATION =',13,//,5X,12E10.3,/) MA 04840 490 FORMAT (/,5X,120('-')) MA 04850 500 FORMAT (/,3X, 'STRESSES IN EACH LAYER, GAUSS POINT AND ELEMENT',/) MA 04860 510 FORMAT (/,lOX, 'ELEMENT NUMBER=', 13,/) MA 04870 520 FORMAT(8Xl 'SIGMAXT' ,5X, 'SIGMAXB' ,5X, 'SIGMAYT' ,5X, 'SIGMAYB' ,5X, 'SIGMA 04880
C ********************~************************************************STI00490 C STI00500 C CALL SUBROUTINE SHAPE TO CALCULATE THE SHAPE FUNCTIONS, THEIR STI00510 C DERIVATIVES, AND THE JACOBIAN MATRIX STI00520 C STI00530 C **********************************************************************STI00540 C STI00550
c
CALL SHAPE (DET,ELXY,ETA,Xl,NPE) CNST=DET*WT(Nl,NGP)*WT(NJ,NGP) X=O.O Y=O.O FX=O.O FY=O.O DO 20 1=1,NPE X=X+SF( I )*ELXY( I, 1) Y=Y+SF( I )*ELXY( 1,2)
C **********************************************************************STI00670 C ST 100680 C CALCULATE THE SINUSOIDAL LOAD ON THE QUARTER PLATE IF ILOAD = 2 STI00690 C ST 100700 c c
CX=Pl*0.5/XL CY=P I *O. 5/Yl P=PO*DCOS(CX*X)*DCOS(CY*Y)
STI00720 STI00730 STI00740 STI00750 STI00760
C **********************************************************************STl00770 C ST I 00780 C CALCULATE THE UNIFORMLY DISTRIBUTED LOAD IF ILOAD IS NOT EQUAL TO 2 STI00790 C ST I 00800 c c
IF ( ILOAD.NE.2) P=PO STI00830 DO 30 1=1,NPE STI00840 L=( l-1)*NDF+1 STI00850 FX= FX+O. 5*CDSF ( 1, I ) *W( L +N3) ST I 00860 FY=FY+0.5*GDSF(2, I )*W(L+N3) STl00870 IF ( ILOAD.EQ.0) GO TO 30 STI00880 ELP(L+N3)=ELP(L+N3)+CNST*SF( I )*P STI00890
30 CONTINUE STI00900 11=1 STI00910 DO 70 1=1,NPE STI00920 JJ=1 STI00930 DO 60 J=1,NPE STI00940 IF (NDF.EQ.3) GO TO 45 STI00950 H(ll,JJ)=H(ll,JJ) + C1*SF(l)*SF(J)*CNST STI00960 H(11+1,JJ+l)=H(ll+1,JJ+1) + C1*SF(l)*SF(J)*CNST STI00970 IF(NT.GT.1 .OR. IT.GT.1 )GOTO 40 STI00980 ST I F( I I , JJ ) =ST I F ( I I , JJ) +( GDSF ( 1, I ) *(A( 1, 1 ) *CDSF ( 1 , J ) +A( 1 , 3 ) *GOS F ( 2ST I 00990
1,J))+GOSF(2, I )*(A(1,3)*GOSF(1,J)+A(3,3)*GOSF(2,J)))*CNST STI01000
c
74
ST I F ( I I , JJ +1 )=ST I F ( I I , JJ+1 ) +( GOSF ( 1 , I ) *(A( 1 , 2) *GOS F ( 2, J ) +A( 1, 3 ) *GOST I 01010 1SF(1,J))+GOSF(2, I )*(A(3,2)*GOSF(2,J)+A(3,3)*GOSF(1,J)))*CNST STI01020
ST I F ( I I +1, JJ ) =ST I F ( I I +1, JJ ) +( GOSF ( 1, J ) *(A( 1 , 2) *GOS F ( 2, I ) +A( 1 , 3 ) *GOST I 01030 1 SF ( 1, I ) ) +GDSF ( 2, J ) *(A( 3, 2) *GOSF ( 2, I ) +A( 3, 3) *GOSF ( 1, I ) ) ) *CNST ST I 01040
STI F( I I ,JJ+N2)=STI F( I I ,JJ+N2)+(GDSF( 1, I )*(A( 1,4)*GOSF( 1,J )+A( 1,6)*STI01050 1GOSF(2,J))+GOSF(2, I )*(A(3,4)*GOSF(1,J)+A(3,6)*GOSF(2,J)) )*CNST STI01060
STI F( 1 l+N2,JJ )=ST IF( I l+N2,JJ )+( GOSF( 1,J )*(A( 1,4)*GOSF( 1, I )+A( 1, 6 )*STIOl070 1GOSF(2, I ))+GOSF(2,J)*(A(3,4)*GOSF(1, I )+A(3,6)*GOSF(2, I )))*CNST STI01080
STIF( 11,JJ+Nl )=STIF( 11,JJ+Nl )+(GDSF( 1, I )*(A( 1,5)*GOSF(2,J)+A( 1,6)*STI01090 1GOSF(1,J))+GOSF(2, I )*(A(3,5)*GOSF(2,J)+A(3,6)*GOSF(1,J)))*CNST STI01100
STIF( I l+N1,JJ)=STIF( I l+N1,JJ)+(GOSF(l,J)*(A(1,5)*GOSF(2, I )+A(1,6)*STI01110 1 GOSF ( 1 , I ) ) +GOSF ( 2, J ) * (A( 3, 5) *GOSF ( 2, I ) +A( 3, 6) *GOSF ( 1 , I ) ) ) *CNST ST I 01120
ST I F ( I I +1, JJ +1 ) =ST I F ( I I +1, JJ+l ) +( GOSF ( 1, I ) * (A( 3, 2) *GOS F ( 2, J) +A( 3, 3STI011 30 1)*GOSF(1,J))+GDSF(2, l)*(A(2,2)*GOSF(2,J)+A(2,3)*GOSF(1,J)))*CNST STI01140
ST I F ( I I +1, JJ +N2 )=ST I F ( I I+ 1 , JJ +N2) +( GOSF ( 1, I ) *(A( 3, 4 l *GOS F ( 1 , J ) +A( 3STI01150 1,6)*GOSF(2,J))+GOSF(2, I )*(A(2,4)*GOSF(1,J)+A(2,6)*GOSF(2,J)))*CNSTSTI01160
STI F( I l+N2,JJ+1 )=STI F( 1 l+N2,JJ+l )+(GOSF( l ,J )*(A( 3,4)*GOSF( 1, I )+A( 3STI01170 1, 6) *GOSF ( 2, I ) ) +GOSF ( 2, J ) *(A( 2, 4) *GOSF ( 1, I ) +A( 2, 6) *GOS F ( 2, I ) ) ) *CNSTST I 01180
ST I F ( I I +1, JJ +N 1 )=ST I F ( I I+ 1, JJ+N 1 ) + ( GOSF ( 1, I ) *(A( 3, 5) *GOSF ( 2, J ) +A( 3STI01190 1,6)*GOSF(1,J))+GOSF(2, I )*(A(2,5)*GOSF(2,J)+A(2,6)*GOSF(1,J)))*CNSTSTI01200
ST I F ( I I +N1, JJ+ 1 )=ST I F ( I I +N 1 , JJ+l ) + (GOS F ( 1 , J ) * (A( 3, 5) *GOS F ( 2, I ) +A( 3STI01210 1, 6) *GOSF ( 1, I ) ) +GOSF ( 2, J ) * (A( 2, 5) *GOSF ( 2, I ) +A( 2, 6) *GOS F ( 1 , I ) ) ) *CNSTST I 01220
GOTO 45 STI01230 STl01240
C **********************************************************************STI01250 C ST I 01260 C CALCULATE THE UNSYMMETRIC STIFFNESS COEFFICIENTS DUE TO THE STI01270 C GEOMETRIC NONLINEARITIES STI01280 C ST 101290 C **********************************************************************STI01300 C STI01310
c
40 CONTINUE STI01320 ST I F ( I I , JJ+N3 ) =ST I F ( I I , JJ +N3) +( GOSF ( 1, I ) * (A( 1, 1 ) * FX*GOS F ( 1, J ) +A( 1, ST I 013 30
12)*FY*GOSF(2,J)+A( 1,3)*(FX*GOSF(2,J)+FY*GOSF(1,J)))+GOSF(2, I )*(A(3STI01340 2,1 )*FX*GOSF(1,J)+A(3,2)*FY*GOSF(2,J)+A(3,3)*(FX*GDSF(2,J)+FY*GDSF(STI01350 31,J))))*CNST STI01360
ST I F ( I I +N3, JJ ) =ST I F ( I I +N3, JJ ) +( GDSF ( 1, J ) *(A( 1 , 1 ) * FX*GDS F( 1 , I ) +A( 1 , ST I 01370 12)*FY*GOSF(2, I )+A(1,3)*(FX*GOSF(2, I )+FY*GOSF(1, I )))+GDSF(2,J)*(A(3STI01380 2,1)*FX*GOSF(1, I )+A(3,2)*FY*GOSF(2, I )+A(3,3)*(FX*GOSF(2, I )+FY*GOSF(STI01390 31,l))))*CNST*2.0 STI01400
STIF( I 1+1,JJ+N3)=STIF( I 1+1,JJ+N3)+(GOSF(2, I )*(A(1,2)*FX*GDSF(1,J)+STI01410 1A(2,2)*FY*GDSF(2,J)+A(2,3)*(FX*GDSF(2,J)+FY*GDSF(1,J)) )+GDSF(1, I )*STl01420 2(A(1,3)*FX*GDSF(1,J)+A(2,3)*FY*GDSF(2,J)+A(3,3)*(FX*GOSF(2,J)+FY*GSTl01430 30SF( 1,J))) )*CNST STI01440
STI F( I l+N3,JJ+1 )=STI F( I l+N3,JJ+1 )+2.0*(GOSF(2,J )*(A( 1,2)*FX*GDSF( 1STI01450 1, I )+A(2,2)*FY*GDSF(2, I )+A(2,3)*(FX*GDSF(2, I )+FY*GDSF( 1, I )))+GDSF(1STI01460 2, J ) * (A( 1 , 3) * FX*GDSF ( 1, I ) +A( 2, 3) *FY*GOSF ( 2, I ) +A( 3, 3) * ( FX*GOSF ( 2, I ) +ST I 014 70 3FY*GOSF(1, I ))J)*CNST STI01480
ST IF( I I +N2, JJ+N2 )=ST IF( I I +N2, JJ+N2) +( GDSF( 1, I ) *(A( 4, 4) *GDSF ( 1, J) +AST I 01510 1(4,6)*GDSF(2,J) )+GDSF(2, I )*(A(4,6)*GOSF(1,J)+A(6,6)*GOSF(2,J)))*CNSTI01520 2ST STI01530
STIF( I l+N2,JJ+N1)=STIF( I l+N2,JJ+N1)+(GDSF(1, I )*(A(4,5)*GDSF(2,J)+ASTI01540 1(4,6)*GDSF(1,J))+GOSF(2, I )*(A(5,6)*GOSF(2,J)+A(6,6)*GOSF(1,J)))*CNSTI01550
75
2ST STI01560 ST IF( I I +N1, JJ+N2 )=ST IF( I I +Nl, JJ+N2) +( GDSF( 1, J) *(A( 4, 5) *GDSF ( 2, I ) +AST I 015 70
1 ( 4, 6) *GOSF ( 1, I ) ) +GOSF ( 2, J ) *(A( 5, 6) *GDSF ( 2, I ) +A( 6, 6) *GDSF ( 1, I ) ) ) *CNST I 01580 2ST STI01590
ST IF ( I I +Nl, JJ+Nl )=ST IF( I I +N1, JJ+Nl )+( GOSF( 1, I ) *(A( 5, 6) *GOSF( 2, J) +AST I 01600 1 ( 6, 6 )*GDSF( 1 ,J) )+GOSF( 2, I)*( A( 5, 5 )*GOSF( 2,J )+A( 5, 6 )*GDSF( 1 ,J)) )*CNST I 01610 2ST STI01620
1FX*GOSF(2,J)+FY*GOSF(1,J)))+FY*GDSF(2, I )*(A(1,2)*FX*CDSF(1,J)+A(2,STI01670 22)*FY*GDSF(2,J)+A(2,3)*(FX*GOSF(2,J)+FY*GOSF(1,J))) STI01680
SW2=(FX*GDSF(2, I )+FY*GOSF(l, I ))*(A(1,3)*FX*GDSF(1,J)+A(2,3)*FY*GDSSTI01690 1 F ( 2, J ) +A ( 3, 3 ) * ( FX*GDS F ( 2, J ) + FY*GDS F ( 1 , J ) ) ) ST I 01700
STIF( I l+N3,JJ+N3)=STIF( I l+N3,JJ+N3)+2.0*(SW1+SW2)*CNST STl01710 ST IF( I I +N3, JJ+N2 )=ST IF( I I +N3, JJ+N2 )+2. 0*( FX*GOSF( 1, I)*( A( 1, 4 )*GOSFST I 01720
1(1,J)+A(1,6)*GOSF(2,J) )+FY*GOSF(2, I )*(A(2,4)*GOSF(1,J)+A(2,6)*GDSFSTI01730 2( 2, J)) +( FY*GOSF ( 1, I ) +FX*GOSF ( 2, I ) ) *(A( 3, 4) *GOSF ( 1, J) +A( 3, 6) *GOSF( 2ST I 01740 3,J)))*CNST STI01750
STIF( I l+N2,JJ+N3)=STIF( I l+N2,JJ+N3)+(FX*GOSF(1,J)*(A( 1,4)*GOSF( 1, ISTI01760 1 )+A(1,6)*GOSF(2, I ))+FY*GOSF(2,J)*(A(2,4)*GOSF(1, I )+A(2,6)*GOSF(2, ISTI01770 2) ) +( FY*GDS F ( 1 , J ) +FX*GDSF ( 2, J) ) *(A( 3, 4) *GOS F ( 1 , I ) +A( 3, 6) *GDSF ( 2, I ) ) ST I 01780 3)*CNST STI01790
ST IF( I I +N3,JJ+N 1 )=ST IF( I I +N3, JJ+Nl )+2. 0*( FX*GDSF( 1, i )*(A( 1, 5 )*GDSFST I 01800 1(2,J)+A(1,6)*GDSF(1,J))+FY*GDSF(2, I )*(A(2,5)*GDSF(2,J)+A(2,6)*GOSFSTI01810 2 ( 1, J ) ) +( FY*GDSF ( 1, I ) +FX*GDSF ( 2, I ) ) *(A( 3, 5) *GOSF ( 2, J ) +A( 3, 6) *GDSF ( 1STI01820 3,J)))*CNST STI01830
ST I F ( I I +N 1, JJ+N3) =ST I F ( I I +N 1 , JJ +N3) +( FX*GDSF ( 1 , J ) * (A( 1 , 5) *GDSF ( 2, I ST I 01840 1 ) +A( 1, 6) *GDSF( 1, I ) ) +FY*GDSF( 2, J) *(A( 2, 5) *GDSF( 2, I ) +A( 2, 6) *GDSF( l, I ST I 01850 2))+(FY*GDSF(1,J)+FX*GDSF(2,J))*(A(3,5)*GOSF(2, I )+A(3,6)*GDSF(1, I ))STI01860 3)*CNST STI01870
C STI01880 55 CONTINUE STI01890
H( I l+N3,JJ+N3)=H( I l+N3,JJ+N3) + Cl*SF( I )*SF(J)*CNST 5TI01900 H( I l+Nl,JJ+Nl )=H( 1 l+Nl,JJ+Nl) + C2*SF( I )*SF(J)*CNST STI01910 H( I l+N2,JJ+N2)=H( I l+N2,JJ+N2) + C2*SF( I )*SF(J)*CNST STI01920
60 JJ=NOF*J+1 STI01930 70 I l=NDF*l+l STI01940 80 CONTINUE STI01950
IF(NT.GT.1 .OR. IT.GT.l)GOTO 130 STI01960 DO 110 Nl=l,LGP STI01970 DO 110 NJ=1,LGP STI01980 Xl=GAUSS(Nl,LGP) STI01990 ETA=GAUSS(NJ,LGP) STl02000 CALL SHAPE (DET,ELXY,ETA,Xl,NPE) STI02010 CNST=CET*WT(NI ,LGP)*WT(NJ,LGP) STI02020 11=1 STI02030 DO 100 1=1,NPE STI02040 JJ=1 STI02050 DO 90 J=l,NPE STI02060 STIF( I l+N3,JJ+N3)=STIF( I l+N3,JJ+N3)+(GOSF(1, I )*(A45*GOSF(2,J)+A55*STI02070
1GOSF(1,J))+GDSF(2, I )*(A44*GDSF(2,J)+A45*GDSF(1,J)))*CNST STI02080 STIF( I l+N3,JJ+N2)=STIF( I l+N3,JJ+N2)+(A55*GDSF(1, I )*SF(J)+A45*GOSF(STl02090
12, I )*SF(J))*CNST STI02i00
c
90 100 110
120 130
140
180 200
76
STIF( I l+N2,JJ+N3)=STIF( I l+N2,JJ+N3)+(A55*GDSF(1,J)*SF( I )+A45*GOSF(STI02110 12,J )*SF( I) )*CNST STl02120
ST IF( I I +N3, JJ+N1 )=ST IF( I I +N3, JJ+Nl )+( A45*GDSF( 1, I ) *SF( J )+A44*GDSF( ST I 02130 12, I )*SF(J))*CNST STI02140
ST IF( I I +N1, JJ+N3 )=ST IF( I I +Nl, JJ+N3 )+( A45*GDSF( 1, J )*SF( I )+A44*GDSF( ST I 02150 12,J)*SF(l))*CNST STI02160
STIF( I l+N2,JJ+N2)=STIF( I l+N2,JJ+N2)+A55*SF( I )*SF(J)*CNST STI02170 ST! F( I l+N2,JJ+Nl )=STI F( I l+N2,JJ+Nl )+A45*SF( I )*SF(J )*CNST STI02180 STIF( I l+N1,JJ+N2)=STIF( I l+N1,JJ+N2)+A45*SF( I )*SF(J)*CNST STI02190 STI F( I l+N1,JJ+Nl )=STI F( I 1+N1 ,JJ+N1 )+A44*SF( I )*SF(J )*CNST STI02200 JJ=NDF*J+1 STI02210 I l=NDF*l+l STI02220 CONTINUE STI02230 DO 120 1=1,NN STI02240 DO 120 J=l,NN STI02250 STFL( N, I ,J )=STI F( I ,J) ST 102260 ST IF( I, J )=0. 0 ST I 02270 CONTINUE STI02280 DO 140 1=1,NN STI02290 DO 140 J=l,NN STI02300 STIF( l,J)=STIF( 1,J)+STFL(N, l,J)+AO*H( l,J) STI02310 DO 200 1=1,NN STI02320 SUM=O.O STI02330 DO 180 K=l,NN STI02340 SUM=SUM+H( I ,K)*(AO*WO(K)+A2*Wl(K)+A3*W2(K)) STI02350 ELP( I )=ELP( I )+SUM ST I 02360 RETURN STI02370 END ST I 02380 SUBROUTINE SHAPE (DET,ELXY,ETA,Xl,NPE) SHA00010
SHA00020 C **********************************************************************SHA00030 C SHA00040 C THIS SUBROUTINE EVALUATES THE SHAPE FUNCTIONS AND THEIR DERIVATIVES SHA00050 C AT THE GUASSIAN POINTS ( ISOPARAMETRIC QUADRILATERAL ELEMENT WITH SHA00060 C NINE NODES). SHA00070 C SHA00080 C ********************************************~*************************SHA00090 c SHAOOlOO
1-1.0D0,2*1.0D0,-1.000,0.0D0,1.0D0,2*0.0DO/ SHA00150 DATA NP/1,2,3,4,5,7,6,8,9/ SHA00160 FNC(A,B)=A*B SHA00170 IF (NPE-8) 50,10,70 SHA00180
10 DO 40 1=1,NPE SHA00190 Nl=NP( I) SHA00200 XP=XNODE(NI, 1) SHA00210 YP=XNODE(Nl,2) SHA00220 XIO=l.O+Xl*XP SHA00230 ETAO=l.O+ETA*YP SHA00240 Xl1=1.0-Xl*XI SHA00250 ETAl=l.0-ETA*ETA SHA00260 IF (I .GT.4) GO TO 20 SHA00270
77
SF{NI )=0.25*FNC(XIO,ETAO)*(Xl*XP+ETA*YP-1.0) DSF(l,NI )=0.25*FNC(ETAO,XP)*(2.0*Xl*XP+ETA*YP) DSF(2,NI )=0.25*FNC(XIO,YP)*(2.0*ETA*YP+Xl*XP) GO TO 40
20 IF ( l.GT.6) GO TO 30 SF(NI )=0.5*FNC(Xl1,ETAO) DSF(l,NI )=-FNC(Xl,ETAO) DSF(2,NI )=0.5*FNC(YP,Xl1) GO TO 40
30 SF(NI )=0.5*FNC(ETA1,XIO) DSF( 1, NI )=O. 5*FNC( XP, ET Al) DSF(2,NI )=-FNC(ETA,XIO)
40 CONTINUE GO TO 120
50 CONTINUE DO 60 1=1,NPE XP=XNODE ( I , 1 ) YP=XNODE( I, 2) XIO=l .O+Xl*XP ETAO=l. O+ETA*YP SF( I )=0.25*FNC(XIO,ETAO) DSF(l, I )=0.25*FNC(XP,ETAO) DSF(2, I )=0.25*FNC(YP,XIO)
60 CONTINUE GO TO 120
70 DO 110 1=1,NPE Nl=NP{I) XP=XNOOE( NI, 1) YP=XNODE( NI, 2) Xl0=1.0+Xl*XP ETAO=l. O+ETA*YP XI 1=1.0-Xl*XI ETA1=1.0-ETA*ETA Xl2=XP*XI ETA2=YP*ETA IF ( l.GT.4) GO TO 80 SF{NI )=0.25*FNC(XIO,ETAO)*Xl2*ETA2 DSF(1,NI )=0.25*XP*FNC(ETA2,ETA0)*{1.0+2.0*Xl2) DSF(2,NI )=0.25*YP*FNC(Xl2,XI0)*(1.0+2.0*ETA2) GO TO 110
80 IF ( l.GT.6) GO TO 90 SF(NI )=0.5*FNC(X11,ETAO)*ETA2 DSF(l,NI )=-Xl*FNC(ETA2,ETAO) DSF( 2, NI )=0. 5*FNC( X 11, YP) *( 1. 0+2. O*ETA2) GO TO 110
90 IF ( l.GT.8) GO TO 100 SF(NI )=0.5*FNC(ETA1,XIO)*Xl2 DSF(2,NI )=-ETA*FNC(Xl2,XIO) DSF( 1, NI )=0. 5*FNC( ETA1 ,XP )*( 1. 0+2. O*X 12) GO TO 110
C **********************************************************************MES00030 C MES00040 C THIS SUBROUTINE GENERATES ARRAY NOD( l,J), COORDINATES X( I ),Y( I), MES00050 C ANO BOUNDARY CONDITIONS FOR RECTANGULAR DOMAINS WHICH ARE MES00060 C DIVIDED INTO LINEAR ( IEL=l) OR QUADRATIC( IEL=2) RECTANGULAR MES00070 C ELEMENTS. COMMON STATEMENT SHOULD MATCH WITH THAT IN MAIN PROGRAM. MES00080 C PLATE BENDING ELEMENTS HAVE THREE DEGREES OF FREEDOM (NDF=3), MES00090 C W,SX,SY, OR FIVE DEGREES OF FREEDOM (NDF=5) U,V,W,SX AND SY. MES00100 c MES00110 C ********************~*************************************************MES00120 C MES00130
C **********************************************************************MES00290 C MES00300 C GENERATE THE ARRAY NOD( I ,J) MES00310 C MES00320 C **********************************************************************MES00330 C MES00340
IF (NPE .EQ. 9)NOD( 1,9)=NXX1+2 IF(NY .EQ. l)GOTO 225 M = 1 DO 220 N = 2,NY L = (N-1 )*NX + 1 DO 210 I = 1,NPE NOD(L, I) = NOD(M, I )+NXXl+( IEL-1 )*NEXl+KO*NX M=L IF(NX .EQ .1 )GO TO 255 DO 250 NI = 2,NX DO 230 I = 1,NPE Kl = IEL IF( I .EQ. 6 .OR. I .EQ. 8)K1=1+KO NOD(NI, I)= NOD(Nl-1, I )+K1 M = NI DO 250 NJ = 2,NY L = (NJ-1 )*NX+NI DO 240 J = 1,NPE NOD(L,J) = NOD(M,J)+NXXl+( IEL-1 )*NEX1+KO*NX M = L
C MES00650 C **********************************************************************MES00660 C MES00670 C GENERATE THE COORDINATES X( I) AND Y( I) MES00680 C MES00690 C **********************************************************************MES00700 C MES00710
255 DO 260 1=1, NXX1 260 DX( I ) = XL/NXX
DO 265 1=1,NYY1 265 DY( I ) = YL/NYY 270 YC=O.O
IF (NPE .EQ. 9) GOTO 300 DO 290 NI = 1, NEY1 I= (NXX1+(1EL-1)*NEX1)*(Nl-1)+1 J = (Nl-1)*1EL+1 X( I) = 0.0 Y( I) = YC DO 275 NJ = 1,NXX 1=1+1 X( I) = X( 1-1 )+DX(NJ)
275 Y( I ) = YC IF(NI .GT. NY .OR. IEL .EQ. 1)GO TO 300 J = J+1 YC = YC+OY(J-1) I = I +1 X( I) = 0.0 Y( I) = YC DO 280 I I = 1, NX K = 2* I 1-1
C **********************************************************************BDU00030 C BDU00040 C EQUATION SOLVER FOR NON-SYMMETRIC SYSTEM OF BDU00050 C EQUATIONS. SOLUTION IS STORED IN A(N,2*1TERM) BDU00060 C BDU00070 c c
C **********************************************************************BDU00170 C BDU00180 C BEGINS ELIMINATION OF THE LOWER LEFT BDU00190 C BDU00200 c c
DO 1000 l=l, N IF (DABS(A( I, !TERM)) .LT. CERO) GO TO 410 GO TO 430 IF (DABS(A(l,ITERM)) .LT. PARE) JLAST =MINO( l+ITERM-1, N) L = I TERM + 1 DO 500 J=I, JLAST L = L - 1
GO TO 1600
IF (DABS(A(J,L)) .LT. PARE) GO TO 500 B = A(J,L) DO 450 K=L, NBND A(J,K) = A(J,K) / B I F ( I . EQ. N ) GO TO 1200 CONTINUE L=O J FIRST = I +
IF (JLAST .LE. I) GO TO 1000 DO 900 J= JFIRST, JLAST L=L+1 IF (DABS(A(J, ITERM-L)) .LT. PARE) GO TO 900 DO 600 K=ITERM, NBM
600 A(J,K-L) = A(J-L,K) - A(J,K-L) A(J,NBND) = A(J-L,NBND) - A(J,NBND) IF (I .GE. N-ITERM+1) GO TO 900 D0800K=1,L
800 900
1000 1200
A(J,NBND-K) = -A(J,NBND-K) CONTINUE CONTINUE L = !TERM - 1 DO 1500 1=2, N DO 1500 J=l, L IF (N+l-i+J .GT. N) GO TO 1500 A(N+l-1,NBNO) = A(N+1-1,NBND) - A(N+1-i+J,NBNO)*A(N+1-I, ITERM+J)
1500 CONTINUE RETURN
1600 WRITE(6,1601) 1601 FORMAT (' COMPUTATION STOPPED IN BNDEQ BECAUSE ZERO APPEARED ON
1MAIN DIAGONAL. THE MATRIX FOLLOWS.') WRITE(6, 1602) I ,A( I, ITERM)
1602 FORMAT(10X, 15,E13.6) STOP END SUBROUTINE MATPRP (LAYER,C,A,H, THETA,AK,G13,G23,A44,A45,A55)
C MAT00020 C **********************************************************************MAT00030 C MAT00040 C SUBROUTINE MATPRP CALCULATES THE STANDARD COMPOSITE MATRICES OBAR MAT00050 C (RELATES STRESS TO STRAINS FOR AN ARBITRARY ORIENTATION), AND MAT00060 C THE A, B, ANO D MATRICES WHICH RELATE THE STRESS ANO MOMENT MAT00070 C RESULTANTS TO THE STRAINS AND ROTATIONS. MAT00080 C MAT00090 C **********************************************************************MAT00100 C MAT00110
C MAT00230 C **********************************************************************MAT00240 C MAT00250 C CALCULATE THE Z'S, DISTANCES TO THE EDGES OF THE LAYERS FROM MAT00260 C THE MIDPLANE MAT00270 C MAT00280 C **********************************************************************MAT00290
c
c
DO 20 1=1,LH 20 ZT=ZT+T( I ) 30 Z(1)=-ZT
DO 40 1=1, LAYER 40 Z( 1+1 )=Z( I )+T( I)
WRITE(6,100)
82
WRITE(6,110) (T(l),1=1,LAYER) WRITE(6,110) Z(l),(Z( 1+1), 1=1,LAYER) WRITE(6,90) DO 50 1=1,6 DO 50 J=1,6
50 A( I ,J )=0.0 A55=0.0 A45=0.0 A44=0.0 DO 70 K=l,LAYER
C **********************************************************************MAT00480 C MAT00490 C READ IN THE ELEMENT PROPERTIES IF LAMINA ARE NOT IDENTICAL MATOOSQO C MAT00510 c c
C **********************************************************************STR00030 C STR00040 C SUBROUTINE STRESS CALCULATES THE STRESS AND BENDING MOMENTS STR00050 C STR00060 c c
DUX=O.O DUY=O.O DVX=O.O DVY=O.O DWX=O.O DWY=O.O DSXY=O.O DSYX=O.O OSXX=O.O DSYY=O.O X=O.O Y=O.O DO 20 1=1,NPE L=( 1-1 )*NDF+l X=X+SF( 1)*ELXY(I,1) Y=Y+SF( I )*ELXY( I, 2) IF (NDF.LT.5) GO TO 10 DUX=DUX+GDSF(l, I )*W(L) DUY=DUY+GDSF(2, l)*W(L) DVX=DVX+GOSF(l, I )*W(L+l) DVY=DVY+GOSF(2, I )*W(L+l)
10 DWX=DWX+GDSF(l, l)*W(L+N3) DWY=DWY+GOSF(2, I )*W(L+N3) SIX=SIX+SF( I )*W(L+N2) SIY=StY+SF( I )*W(L+Nl) DSXX=DSXX+GDSF(l, I )*W(L+N2) DSXY=DSXY+GOSF(2, l)*W(L+N2) DSYX=OSYX+GDSF(l, I )*W(L+Nl)
20 OSYY=DSYY+GOSF(2, I )*W(L+Nl) IF( ITER .EQ. l)GOTO 25 EX = DUX+0.5*DWX*DWX EY = OVY+0.5*0WY*OWY EXY=OUY+DVX+DWX*DWY GOTO 26